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<body class="ignore-math"><article class="example example-like"><p><dfn class="terminology">Example 2</dfn> Find the solution of the following initial value problem:</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_9.html ./knowl/eq2_9.html ./knowl/GPsolution.html">
\begin{equation}
x^3 y^{\prime}+4 x^2 y=e^{-x}, \quad y(-1)=0.\tag{2.1.14}
\end{equation}
</div>
<p class="continuation"><dfn class="terminology">Solution</dfn>: The equation is first transformed to</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_9.html ./knowl/eq2_9.html ./knowl/GPsolution.html">
\begin{equation*}
y^{\prime}+\frac{4}{x}y=\frac{e^{-x}}{x^3}.
\end{equation*}
</div>
<p class="continuation">Then one can see <span class="process-math">\(p(x)=\frac{4}{x}\)</span> and the integrating factor is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_9.html ./knowl/eq2_9.html ./knowl/GPsolution.html">
\begin{equation*}
u(x)=\exp (\int p(x) \textrm{d} x)=\exp (\int \frac{4}{x} \textrm{d} x)=\exp (4 \ln x)=x^4.
\end{equation*}
</div>
<p class="continuation">Multiplying (<a href="" class="xref" data-knowl="./knowl/eq2_9.html" title="Equation 2.1.9">(2.1.9)</a>) with the integrating factor, one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_9.html ./knowl/eq2_9.html ./knowl/GPsolution.html">
\begin{equation*}
\begin{aligned}
&amp;x^4 y^{\prime}+4 x^3 y=x e^{-x} \rightarrow \frac{\textrm{d}}{\textrm{d} x} (x^4 y)=x e^{-x} \rightarrow x^4 y=-x e^{-x}-e^{-x}+C.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Therefore, the general solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_9.html ./knowl/eq2_9.html ./knowl/GPsolution.html">
\begin{equation*}
y=\frac{-x e^{-x}-e^{-x}+C}{x^4}.
\end{equation*}
</div>
<p class="continuation">Applying the initial condition <span class="process-math">\(y(-1)=0\text{,}\)</span> one has <span class="process-math">\(C=0\text{.}\)</span> Therefore, the solution to the initial value problem (<a href="" class="xref" data-knowl="./knowl/eq2_9.html" title="Equation 2.1.9">(2.1.9)</a>) is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_9.html ./knowl/eq2_9.html ./knowl/GPsolution.html">
\begin{equation*}
y=\frac{-x e^{-x}-e^{-x}}{x^4}.
\end{equation*}
</div>
<p class="continuation"><a href="" class="xref" data-knowl="./knowl/GPsolution.html" title="Figure 2.1.3">Figure 2.1.3</a> shows the general solution and the particular solution.</p>
<figure class="figure figure-like"><div class="image-box" style="width: 100%; margin-left: 0%; margin-right: 0%;"><img src="external/./img/GPsolution.eps" class="contained" alt="The general solution and particular solution to Example 2."></div>
<figcaption><span class="type">Figure</span><span class="space"> </span><span class="codenumber">2.1.3<span class="period">.</span></span><span class="space"> </span>The general solution and particular solution to Example 2.</figcaption></figure></article></body>
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